What is the limit of $[x^2-5x+6]/ [x^3-8]$ as x goes to infinity?

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Answer 1 · 2015-10-16T20:17:47

$0$

At infinity, the $x^3$ term in the denominator dominates the function as it grows much quicker than all the other terms.

Therefore we may write :

$lim_(x->oo)(x^2-5x+6)/(x^3-8)=lim_(x->oo)x^2/x^3=lim_(x->oo)1/x=0$

What is the limit of [x^2-5x+6]/ [x^3-8] [x^2-5x+6]/ [x^3-8] as x goes to infinity?